Question

$(3k-1,k-2),(k,k-4),(k-1,k-2)$ are collinear, Find $k$.

Answer 1

Given,

$(3k-1,k-2),(k,k-4),(k-1,k-2)$

Since given points are collinear,

$\Rightarrow Ar.△=0$

$Ar.△=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=0$

$\frac{1}{2}\left[\right(3k-1\left)\right(k-4-k+2)+k(k-2-k+2)+(k-1\left)\right(k-2-k+4\left)\right]=0$

$\frac{1}{2}\left[\right(3k-1\left)\right(-2)+k(0)+(k-1\left)\right(2\left)\right]=0$

$\frac{1}{2}[-6k+2+2k-2]=0$

$\frac{1}{2}[-4k]=0$

$-2k=0$

$\therefore k=0$